What is the resultant #pH# of the solution.....?

What is the resultant #pH# of the solution prepared by taking a #10*mL# volume of #HCl(aq)# of #0.10*mol*L^-1# concentration, and diluting this to a #1*L# volume?

1 Answer
anor277
Apr 21, 2017

#pH=3#

Explanation:

#pH=-log_(10)[H_3O^+]#

#[H_3O^+]=(10xx10^-3*Lxx0.1*mol*L^-1)/(1*L)#

#10^-3*mol*L^-1#

And thus #pH=-log_(10)[H_3O^+]#

#-log_(10)(10^-3)=3...........#

And I did this all without a calculator. Note that normally, we would always add acid to water, and not water to acid. However, here the acid concentration is sufficiently dilute to get away with reverse addition.

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